Show that a relation is a equivalence relation

Question

I have a infinite set $A$, and $F$ is the set of all functions $g \colon A \to A$. Let the equivalence relation $\sim$ on $F$ be defined such that $f \sim g$ if only if the set $D_{fg} = \{ a \in A | f(a) \neq g(a)\}$ is finite. I want to show that $\sim$ really is an equivalence relation. Thanks.

Answer 1

It is direct that the relation is reflexive and symmetric.

Hint on transitivity:

$$D_{fh}\subseteq D_{fg}\cup D_{gh}$$

Answer 2

If $A$ is a measurable space with some conveniently defined $\mu$ on it, your binary relation can be turned into an equivalence $f\sim g$ in the sense $f$ and $g$ are equal "almost everywhere", that is to say everywhere but on a set $D_{fg}$ of measure zero. $\mu(D_{fg})=0$. This is commonly done in the theory of integration. But you have to ship $A$ with a $\sigma$-additive measure of sets.

The relation $R(f,g)$: the set $D_{fg}=\{x\in A, f(x)\neq g(x)\}$ is negligible (has measure zero) is an equivalence relation (symmetric, reflexive, and transitive) not only because,

$D_{fh}=\{x\in A, f(x)\neq h(x)\}\subset D_{fg}=\{x\in A, f(x)\neq g(x)\} \cup D_{gh}=\{x\in A, g(x)\neq h(x)\}$

but conversely because the union of two negligible sets is a negligible set. The concept of $\sigma$-additive set measurability is at the heart of the proof.